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Question: Express the complex number \[\dfrac{3+2i}{-2+i}\] in the standard form of \[a+ib\]....

Express the complex number 3+2i−2+i\dfrac{3+2i}{-2+i} in the standard form of a+iba+ib.

Explanation

Solution

Hint: To solve this type of problem we have to write the rationalizing term and multiply it with both numerator and denominator. After rationalizing, solve the expression further to get by using standard formulas to get the solution.

Complete step-by-step answer:
Given 3+2i−2+i\dfrac{3+2i}{-2+i}
We have to multiply 3+2i−2+i\dfrac{3+2i}{-2+i} with −2−i-2-i on both numerators and denominators.
The expression appears as follows,
3+2i−2+i×−2−i−2−i\dfrac{3+2i}{-2+i}\times \dfrac{-2-i}{-2-i}
By solving further we get,
−6−4i−3i−2i24−i2\dfrac{-6-4i-3i-2{{i}^{2}}}{4-{{i}^{2}}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
We know that i2{{i}^{2}}= -1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (b)
Now by substituting (b) in (a) we get,
−6−4i−3i−2(−1)4−(−1)\dfrac{-6-4i-3i-2(-1)}{4-(-1)}
Further solving the above expression we get,
−6+2−7i5\dfrac{-6+2-7i}{5}
Further solving the above expression we get,
−4−7i5\dfrac{-4-7i}{5}
=−45−75i=\dfrac{-4}{5}-\dfrac{7}{5}i which is in the form of a+iba+ib.

Note: This is a direct problem which just requires the rationalizing of the expression. The value of a and b can be any number either rational, whole etc. Therefore while writing the definition of complex numbers we write a,b∈Ra,b\in R.